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1 PROGRAM OF “PHYSICS” Lecturer: Dr. DO Xuan Hoi Room 413
2PHYSICS 2 (FLUID MECHANICS AND THERMAL PHYSICS) 02 credits (30 periods)Chapter 1 Fluid MechanicsChapter 2 Heat, Temperature and the Zeroth Law of ThermodynamicsChapter 3 Heat, Work and the First Law of ThermodynamicsChapter 4 The Kinetic Theory of GasesChapter 5 Entropy and the Second Law ofThermodynamics
3References :Halliday D., Resnick R. and Walker, J. (2005), Fundamentals of Physics, Extended seventh edition. John Willey and Sons, Inc.Alonso M. and Finn E.J. (1992). Physics, Addison-Wesley Publishing CompanyHecht, E. (2000). Physics. Calculus, Second Edition. Brooks/Cole.Faughn/Serway (2006), Serway’s College Physics, Brooks/Cole.Roger Muncaster (1994), A-Level Physics, Stanley Thornes.
5Entropy and the Second Law Chapter 5Entropy and the Second Lawof Thermodynamics1 Heat Engines andthe Second Law of Thermodynamics2 Reversible and Irreversible Processes3 The Carnot Engine4 Gasoline and Diesel Engines5 Heat Pumps and Refrigerators6 Entropy
61 Heat Engines andthe Second Law of Thermodynamics1.1 NotionA heat engine is a device that converts internal energy (heat) to other useful forms,such as electrical or mechanical energyA heat engine carries some working substance through a cyclical processIn internal-combustion engines, such as those used in automobiles, the working substance is a mixture of air and fuel; in a steam turbine it is water.
71.2 Working principle of a heat engine Energy is transferred from a source at a high temperature (Qh)Work is done by the engine (Weng)Energy is expelled to a source at a lower temperature (Qc)Schematic representationof a heat engine
8Since it is a cyclical process, U = 0 Its initial and final internal energies are the sameTherefore, from the first law of thermodynamics :U = Qnet – Weng = 0 Qnet = Weng ; Qh – Qc = WThe work done by the engine equals the net energy absorbed by the engineThe work is equal to the area enclosed by the curve of the PV diagram
91.3 Thermal Efficiency of a Heat Engine Thermal efficiency is defined as the ratio of the work done by the engine to the energy absorbed at the higher temperaturee = 1 (100% efficiency) only if Qc = 0No energy expelled to cold reservoir
101.4 Second Law of Thermodynamics “It is impossible to construct a heat engine that, operating in a cycle, produces no other effect than the absorption of energy from a reservoir and the performance of an equal amount of work”Means that Qc cannot equal 0Some Qc must be expelled to the environmentMeans that e cannot equal 100%The impossible engine
11Find the efficiency of a heat engine that absorbs J of energy from a hot reservoir and exhausts1500 J to a cold reservoir.PROBLEM 1SOLUTION
12A diesel engine performs 2200 J of mechanical work and discards 4300 J of heat each cycle.(a) How much heat must be supplied to the engine in eachcycle?(b) What is the thermal efficiency of the engine?PROBLEM 2SOLUTION
13An aircraft engine takes in 9000 J of heat and discards 6400 J each cycle.(a) What is the mechanical work output of the engine during one cycle?(b) What is the thermal efficiency of the engine?PROBLEM 3SOLUTION
14A gasoline engine has a power output of 180 kW (about 241 hp). Its thermal efficiency is 28.0%.(a) How much heat must be supplied to the engine per second?(b) How much heat is discarded by the engine per second?PROBLEM 4SOLUTION
15A gasoline engine in a large truck takes in 10,000 J of heat and delivers 2000 J of mechanical work per cycle. The heat isobtained by burning gasoline with heat of combustionLC = 5.0 104 J/g.(a) What is the thermal efficiency of this engine?(b) How much heat is discarded in each cycle?(c) How much gasoline is burned in each cycle?(d) If the engine goes through 25 cycles per second, what is itspower output in watts? In horsepower?PROBLEM 5SOLUTION(a)(b)
16A gasoline engine in a large truck takes in 10,000 J of heat and delivers 2000 J of mechanical work per cycle. The heat isobtained by burning gasoline with heat of combustionLC = 5.0 104 J/g.(a) What is the thermal efficiency of this engine?(b) How much heat is discarded in each cycle?(c) How much gasoline is burned in each cycle?(d) If the engine goes through 25 cycles per second, what is itspower output in watts? In horsepower?PROBLEM 5SOLUTION(c)(d)
172 Reversible and Irreversible Processes Thermodynamic processes that occur in nature are all irreversible processes.These are processes that proceed spontaneouslyin one direction but not the otherEXAMPLE : The flow of heat from a hot body to a cooler body Sliding a book across a table converts mechanical energy into heat by friction(a book initially at rest on the table would spontaneously start moving and the table and book would cool down ?). A block of ice melts irreversibly when we place it in a hot (70°C) metal box.
182 Reversible and Irreversible Processes A reversible process is one in which every state along some path is an equilibrium stateAnd one for which the system can be returned to its initial state along the same pathAn irreversible process does not meet these requirementsMost natural processes are irreversibleReversible process are an idealization, but some real processes are good approximationsReversible processes are thus equilibrium processes, with the system always in thermodynamic equilibrium.
193 THE CARNOT ENGINE A theoretical engine developed by Sadi Carnot A heat engine operating in an ideal, reversible cycle (now called a Carnot Cycle) between two reservoirs is the most efficient engine possibleCarnot’s Theorem : No real engine operating between two energy reservoirs can be more efficient than a Carnot engine operating between the same two reservoirs
20Carnot Cycle, A to B A to B is an isothermal expansion The gas is placed in contact with the high temperature reservoirThe gas absorbs heat QhThe gas does work WAB in raising the piston
21Carnot Cycle, B to C B to C is an adiabatic expansion The base of the cylinder is replaced by a thermally nonconducting wallNo heat enters or leaves the systemThe temperature falls from Th to TcThe gas does work WBC
22Carnot Cycle, C to DThe gas is placed in contact with the cold temperature reservoirC to D is an isothermal compressionThe gas expels energy QCWork WCD is done on the gas
23Carnot Cycle, D to A D to A is an adiabatic compression The gas is again placed against a thermally nonconducting wallSo no heat is exchanged with the surroundingsThe temperature of the gas increases from TC to ThThe work done on the gas is WCD
24 The Camot cycle consists of two reversible isothermal and two reversible adiabatic processes.
25Carnot Cycle, PV Diagram The work done by the engine is shown by the area enclosed by the curveThe net work is equal to Qh - QcThe thermal efficiencyof the engine :We can demonstrate : The thermal efficiency of a Carnot engine :(The efficiency of a Carnot engine depends only on thetemperatures of the two heat reservoirs)
26Notes About Carnot Efficiency Efficiency is 0 if Th = TcEfficiency is 100% only if Tc = 0 KSuch reservoirs are not availableThe efficiency increases at Tc is lowered and as Th is raisedIn most practical cases, Tc is near room temperature, 300 KSo generally Th is raised to increase efficiencyAll real engines are less efficient than the Carnot engineReal engines are irreversible because of frictionReal engines are irreversible because they complete cycles in short amounts of time
27A steam engine has a boiler that operates at 500 K. The energy from the burning fuel changes water tosteam, and this steam then drives a piston. The coldreservoir’s temperature is that of the outside air,approximately 300 K. What is the maximum thermal efficiencyof this steam engine?PROBLEM 6SOLUTION
28The highest theoretical efficiency of a certain engine is 30%.If this engine uses the atmosphere, which has a temperatureof 300 K, as its cold reservoir, what is the temperature of itshot reservoir?PROBLEM 7SOLUTION
29A Carnot engine takes 2000 J of heat from a reservoir at 500 K, does some work, and discards some heatto a reservoir at 350 K. How much work does it do, how muchheat is discarded, and what is the efficiency ?PROBLEM 8SOLUTION(Qc < 0 : heat flows out of the engine and into the cold reservoir.)
30Suppose 0.200 mol of an ideal diatomic gas ( = 1.40) undergoes a Carnot cycle with temperatures 227°Cand 27°C. The initial pressure is Pa = 10.0 105 Pa, andduring the isothermal expansion at the higher temperaturethe volume doubles. (a) Find the pressure and volume at eachof points a, b, c, and d in the P V-diagram of the figure.PROBLEM 9SOLUTION
31Suppose 0.200 mol of an ideal diatomic gas ( = 1.40) undergoes a Carnot cycle with temperatures 227°Cand 27°C. The initial pressure is Pa = 10.0 105 Pa, andduring the isothermal expansion at the higher temperaturethe volume doubles. (a) Find the pressure and volume at eachof points a, b, c, and d in the P V-diagram of the figure.PROBLEM 9SOLUTION
32Suppose 0.200 mol of an ideal diatomic gas ( = 1.40) undergoes a Carnot cycle with temperatures 227°Cand 27°C. The initial pressure is Pa = 10.0 105 Pa, andduring the isothermal expansion at the higher temperaturethe volume doubles. (b) Find Q, W, and U for each stepand for the entire cycle.PROBLEM 9SOLUTION
33Suppose 0.200 mol of an ideal diatomic gas ( = 1.40) undergoes a Carnot cycle with temperatures 227°Cand 27°C. The initial pressure is Pa = 10.0 105 Pa, andduring the isothermal expansion at the higher temperaturethe volume doubles. (b) Find Q, W, and U for each stepand for the entire cycle.PROBLEM 9SOLUTION
34Suppose 0.200 mol of an ideal diatomic gas ( = 1.40) undergoes a Carnot cycle with temperatures 227°Cand 27°C. The initial pressure is Pa = 10.0 105 Pa, andduring the isothermal expansion at the higher temperaturethe volume doubles. (b) Find Q, W, and U for each stepand for the entire cycle.PROBLEM 9SOLUTION
35Suppose 0.200 mol of an ideal diatomic gas ( = 1.40) undergoes a Carnot cycle with temperatures 227°Cand 27°C. The initial pressure is Pa = 10.0 105 Pa, andduring the isothermal expansion at the higher temperaturethe volume doubles. (c) Determine the efficiency directly fromthe results of part (b).PROBLEM 9SOLUTION
364 Internal-Combustion Engines Gasoline and Diesel Engines(used in automobiles and many other types of machinery) are familiar examples of a heat engine.a. Gasoline Engines : In a gasoline engine, six processes occur in each cycle
37Cycle of a four-stroke internal-combustion engine. Intake stroke: Piston moves down, causing a partial vacuum in cylinder; gasoline-air mixture enters through intake valve.
38Cycle of a four-stroke internal-combustion engine. Compression stroke: Intake valve closes; mixture is compressed as piston moves up.Ignition: Spark plug ignites mixture.Power stroke: Hot burned mixture expands, pushing piston down.
39Cycle of a four-stroke internal-combustion engine. Exhaust stroke: Exhaust valve opens; piston moves up, expelling exhaust and leaving cylinder ready for next intake stroke.
40Cycle of a four-stroke internal-combustion engine.
41The Otto cycleThe Otto cycle is an idealized model of the thermodynamicprocesses in a gasoline engineIf the air–fuel mixture is assumed to be an ideal gas, then the efficiency of the Otto cycle iswhere is the ratio of the molar specific heats :and V1/V2 is the compression ratioNOTES : For a typical compression ratio of 8 and with = 1.4, we predict a theoretical efficiency of 56% Efficiencies of real gasoline engines are typically around 35%
42b. The Diesel Cycle Diesel engines operate on a cycle similar to the Otto cycle but do not employ a spark plug. There is no fuel in the cylinder at the beginning of the compression stroke. The compression ratio for a diesel engine is much greater than that for a gasoline engine Diesel engines are more efficient than gasoline engines because of their greater compression ratios and resulting higher combustion temperatures.
43What compression ratio r must an Otto cycle have to achieve an ideal efficiency of 65.0% if = 1.40 ?PROBLEM 10SOLUTIONNOTES : e increases when r increases.
44(a) Calculate the theoretical efficiency for an Otto cycle engine with = 1.40 and r = 9.50.(b) If this engine takes in 10,000 J of heat from burning itsfuel, how much heat does it discard to the outside air ?PROBLEM 11SOLUTION(a)(b)
455 Heat Pumps and Refrigerators In a refrigerator or heat pump, theengine absorbs energy Qc from a cold reservoir and expels energy Qh to a hot reservoirsome work must be done on the engineRefrigeratorImpossible refrigerator
46 The principles of the common refrigeration cycle The fluid "circuit" contains a refrigerant fluid (the working substance)The compressor takes in fluid, compresses it adiabatically, and delivers it to the condenser coil at high pressure.the refrigerant gives off heat QH and partially condenses to liquidThe fluid then expands adiabatically into the evaporatorAs the fluid expands, it cools considerably, enough that the fluid in the evaporator coil is colder than its surroundings.
47 Effectiveness of a refrigerator: coefficient of performance (COP) QC > W COP >1It is desirable for the COP to be as high as possibleRefrigeratorTo compare with the thermal efficiency of a Heat Engine:
48A refrigerator has a coefficient of performance of In each cycle it absorbs 3.40104 J ofheat from the cold reservoir.(a) How much mechanical energy is required each cycle tooperate the refrigerator?(b) During each cycle, how much heat is discarded to thehigh-temperature reservoir?PROBLEM 12SOLUTION
496 Entropy Pressure and volume are state properties, properties that depend only on the state of the gas and not on how itreached that state. Other state properties are temperature and energy. Another state property: its entropy. Entropy S of the system is defined by :J/K(The change in entropy of a system during an arbitraryprocess between an initial state i and a final state f )
50 To find the entropy change for an irreversible process occurring in a closed system, replace that process with any reversible process that connects the same initial and final states.The isothermal expansion:To keep the temperature T of the gas constant during the isothermal expansion: heat Q must have been energy transferred from the reservoir to the gas Q is positive and the entropy of the gas increases during the isothermal process and during the free expansion
51 Entropy as state function To make the process reversible, it is done slowly in a series of small steps, with the gas in an equilibrium state at the end of each step. For each small step, the energy transferred as heat to or from the gas is dQ, the work done by the gas is dW, and the change in internal energy is dEint.The first law of thermodynamics:
52 The second law of thermodynamics : “The total entropy of an isolated system that undergoes a change can never decrease” The entropy of the Universe increases in all real processes Entropy provides a quantitative measure of disorder.
53ExampleConsider the changes in entropy that occur in a Carnot heat engine operating between the temperatures Tc and Th .But for Carnot cycle :
54 Consider a system taken through an arbitrary (non-Carnot) reversible cycle Entropy S is a state function S depends only onthe properties of a given equilibrium state S = 0 When a system is in a particular macroscopic state, the particles that make up the system may be in any of w possible microscopic states.From a microscopic viewpoint, entropy is defined as :k : Boltzmann’s constant;W : the number of microstates available to the system
55One kilogram of ice at O°C is melted and converted to water at O°C. Compute its change in entropy,assuming that the melting is done reversibly. The heat offusion of water is Lf = 3.34 105 J/kg.PROBLEM 13SOLUTIONThe heat needed to melt the ice :
56Suppose 1.0 mol of nitrogen gas is confined to the left side of the container. You open the stopcock, andthe volume of the gas doubles.What is the entropy change of the gas for this irreversibleprocess? Treat the gas as ideal.PROBLEM 14SOLUTIONThe entropy change for the irreversible process by calculating it for a reversible process that provides the same change in volume. The temperature of the gas does not change isothermal expansion
57One kilogram of water at 0°C is heated to 100°C. Compute its change in entropy.Specific heat of water is 4190 J/kg.KPROBLEM 15SOLUTION
58Suppose 1.00 kg of water at 100°C is placed in thermal contact with 1.00 kg of water at 0°C. What is thetotal change in entropy? Assume that the specific heat ofwater is constant at 4190 J/kg.K over this temperature range.PROBLEM 16SOLUTION
594 Microscopic Interpretation of Entropy 4.1 Macroscopic and microscopic states Toss 4 identical coins on the floor:There can be many microscopic states that correspond to the same macroscopic description.N coins that are all heads constitute a completely ordered macroscopic state(macrostate) The macroscopic description "half heads, half tails" : the system is disordered The state "half heads, half tails"has a much greater number of possible microscopicstates (microstates), much greater disorder :It is the most probable
604.2 Microscopic expression for entropy If entropy is a quantitative measure of disorderFor any system, the most probable macrostate is the one with the greatest number of corresponding microstates, which is also the macrostate with the greatest disorder and the greatest entropy. Let w represent the number of possible microscopic states(or multiplicity) for a given macrostate.the entropy of a macroscopic state :
61 process that takes it from macrostate 1, for which Consider a system that undergoes a thermodynamicprocess that takes it from macrostate 1, for whichthere are w1 possible microstates (multiplicity w1), tomacroscopic state 2, with w2 associated microstates(multiplicity w2) .The change in entropy in this process isThe difference in entropy between the two macrostates depends on the ratio of the numbers of possiblemicrostates.
62between the two halves of a box. For a system of N molecules that may be distributedbetween the two halves of a box.n1: the number of molecules in one half of the box andn2 : the number in the other half.The multiplicity:When N is very large (the usual case):(Stirling's approximation)
63Suppose that there are 1-00 indistinguishable molecules in the box.How many microstates are associated with the configuration(n1 = 50; n2 = 50),and with the configuration (n1 = 100; n2 = 0)?PROBLEM 17SOLUTION Configuration (n1 = 50; n2 = 50)The multiplicity: Configuration (n1 = 100; n2 = 0)The multiplicity:
64 Initial configuration (n1 = 100; n2 = 0) A thermally insulated box is divided by a partitioninto two compartments, each having volume V. Initially, one com-partment contains n moles of an ideal gas at temperature T, andthe other compartment is evacuated. We then break the partition,and the gas expands to fill both compartments. What is the entropychange in this free-expansion process?PROBLEM 18SOLUTIONNB: Use of the equation Initial configuration (n1 = 100; n2 = 0)Initial multiplicity:Initial entropy:
65 Final configuration (n1 = N/2; n2 = N/2) A thermally insulated box is divided by a partitioninto two compartments, each having volume V. Initially, one com-partment contains n moles of an ideal gas at temperature T, andthe other compartment is evacuated. We then break the partition,and the gas expands to fill both compartments. What is the entropychange in this free-expansion process?PROBLEM 18SOLUTION Final configuration (n1 = N/2; n2 = N/2)Final multiplicity:Final entropy:The change in entropy:
66A thermally insulated box is divided by a partition into two compartments, each having volume V. Initially, one com-partment contains n moles of an ideal gas at temperature T, andthe other compartment is evacuated. We then break the partition,and the gas expands to fill both compartments. What is the entropychange in this free-expansion process?PROBLEM 18SOLUTIONWhen the partition is broken, each molecule nowhas twice as much volume in which it can moveand hence has twice the number of possible positions.Let w1 be the number of microscopic states of the system as a whole when the gas occupies volume VThe number w2 of microscopic states when the gas occupies volume 2V is greater by a factor of 2N : w2 = 2Nw1
67V. The gas expands isothermally and reversibly to a volume PROBLEM 19Two moles of an ideal gas occupy a volumeV. The gas expands isothermally and reversibly to a volume3V. (a) Is the velocity distribution changed by the isothermalexpansion? (b) Calculate the change in entropy of the gas.SOLUTION(a) The velocity distribution depends only on T, so in an isothermal process it does not change.(b) The number of possible positions available to each molecule isaltered by a factor of 3 (becomes larger).Hence the number of microscopic states the gas occupies atvolume 3V is w2= (3)Nw1, where N is the number of moleculesand w1 is the number of possible microscopic states at the startof the process.